I find it difficult to take the derivative of something that has a constant base with a variable as an exponent...and simplify it... PLEASE HELP ME!
f(x) = e^(x^(-1))
f'(x) = -1(e^(x^(-2))e^(x^(-1))
f'(x) = -e^(x^(-2))e^(x^(-1))
?????
and... f(x) = 3^x
f'(x) = 3^x
d/dx e^u = e^u du/dx
here u = x^-1
du/dx = -1 x^-2
so
d/dx e^(x^-1) = -x^-2 e^(x^-1)
= -(1/x^2)e^(1/x^2)
Try the others using that recipe:
d/dx e^u = e^u du/dx
WOW that actually helps a lot and less confusing to solve! Thanks so much!
To take the derivative of a function with a constant base and a variable as an exponent, you can use the chain rule.
Let's start with the first example: f(x) = e^(x^(-1)).
To find the derivative of this function, we need to apply the chain rule. The chain rule states that if we have a composition of functions, such as f(g(x)), then the derivative of this composition is given by the derivative of the outer function times the derivative of the inner function.
In this case, our composition is f(g(x)) where g(x) = x^(-1) and f(x) = e^x.
To find the derivative of f(g(x)), we need to find the derivative of f(x) with respect to x, and evaluate it at g(x), and then multiply it by the derivative of g(x) with respect to x.
The derivative of f(x) = e^x is simply e^x.
The derivative of g(x) = x^(-1) can be found using the power rule for derivatives. The power rule states that if we have a function of the form f(x) = x^n, then the derivative of f(x) is given by n*x^(n-1).
Applying the power rule, the derivative of g(x) = x^(-1) is (-1)*x^(-2), or -x^(-2).
Now, we can find the derivative of f(g(x)) by evaluating the derivative of f(x) at g(x) and multiplying it by the derivative of g(x).
The derivative of f(g(x)) is e^(g(x))*(-x^(-2)).
Plugging in g(x) = x^(-1), we get f'(x) = e^(x^(-1))*(-x^(-2)).
This is the exact derivative of f(x) = e^(x^(-1)), and it cannot be further simplified.
For the second example: f(x) = 3^x.
To find the derivative of this function, we can again use the chain rule.
The base 3 in this case is a constant, so the derivative of 3^x is simply 3^x * ln(3), where ln(3) is the natural logarithm of 3.
Therefore, f'(x) = 3^x * ln(3).
This is the derivative of f(x) = 3^x, and it does not require further simplification.